Question

In Problems 51–60, write out each sum.

$\sum _{k=0}^{n-1}\frac{1}{3^{k+1}}$

Short answer

The required sum is$\sum _{k=0}^{n-1}\frac{1}{3^{k+1}}=\left\{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{3^n}\right\}$.

Step-by-step solution

Step 1. Write the given information.

The given summation notation is:

$\sum _{k=0}^{n-1}\frac{1}{3^{k+1}}$

Step 2. Write out the sum.

To get the sum, we need to substitute $0,1,2,........,n-1$for k,in the given sequence$\sum _{k=0}^{n-1}\frac{1}{3^{k+1}}$,

$$\begin{gathered} \sum _{k=0}^{n-1}\frac{1}{3^{k+1}}=\left\{\frac{1}{3^{0+1}}+\frac{1}{3^{1+1}}+\frac{1}{3^{2+1}}+......+\frac{1}{3^{n-1+1}}\right\} \\ =\left\{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^n}\right\} \\ =\left\{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{3^n}\right\} \end{gathered}$$

Therefore,$\sum _{k=0}^{n-1}\frac{1}{3^{k+1}}=\left\{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{3^n}\right\}$.